Bilinear Form & Linear Functional: Symmetric & Coercive?

In summary, the bilinear form for the problem -\Deltau + b . \nablau + cu = f(x) in \Omega is B(u,v) = \int_D a1\nablau . \nablav dx + \int_D a0uv dx, and the linear functional is L(v) = \int_D fv dx. The bilinear form is symmetric, meaning B(u,v) = B(v,u) for all u and v. To determine if it is coercive, one needs to check if the function b is bounded below by a constant and use Poincare's inequality. To find the bilinear form, one multiplies the equation by a test function, integrates over the domain,
  • #1
squenshl
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4

Homework Statement


The bilinear form are symmetric, i.e. a(u,v) = a(v,u) for all u and v. Find the bilinear form and the linear functional for the problem -[tex]\Delta[/tex]u + b . [tex]\nabla[/tex]u + cu = f(x) in [tex]\Omega[/tex]
u = 0 on the boundary.
Is this bilinear form for this problem symmteric? Is it coersive (assume C > 0)?


Homework Equations





The Attempt at a Solution

 
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  • #2
?? Why have you not shown any work? What is the "bilinear form" associated with this problem?
 
  • #3
a(u,v) = [tex]\int_D[/tex] a1[tex]\nabla[/tex]u . [tex]\nabla[/tex]v dx + [tex]\int_D[/tex] a0uv dx
 
  • #4
To start off for this problem do we multiply by v, integrate over D and use the divergence theorem?
So we get (-[tex]\Delta[/tex]u + b . [tex]\nabla[/tex]u + cu - f)v dx = 0 in D which implies [tex]\int_D[/tex] (-[tex]\Delta[/tex]u + b . [tex]\nabla[/tex]u + cu)v dx = [tex]\int_D[/tex] fv dx which implies some integral + some integral + [tex]\int_D[/tex] cuv dx = [tex]\int_D[/tex] fv dx, I just don't know how to find those integrals using the divergence theorem. Do i change that -[tex]\Delta[/tex]u to -[tex]\nabla[/tex].[tex]\nabla[/tex]u and would this make it easier?
 
  • #5
I guess what I'm asking is how to find [tex]\int_D[/tex] -[tex]\nabla[/tex] . [tex]\nabla[/tex]u + b . [tex]\nabla[/tex]u dx, b and c are function of x.
 
  • #6
In order to get the bilinear form you multiply the equation by a test function (in this case v), then you integrate over the domain Omega, then you integrate by parts (Note: you always integrate by parts when formulating a weak statement for a PDE .. the entire point is to transfer derivatives onto the test function, thus requiring less regularity), then you can apply divergence theorem to get rid of boundary terms. The coercivity can be tricky unless the function b is bounded below by a constant, but you can always use Poincare's inequality.
 
  • #7
Thanks.
Now all I need to do is just evaluate the integral [tex]\int_D[/tex] -[tex]\nabla[/tex] . [tex]\nabla[/tex]u + b . [tex]\nabla[/tex]u dx
 
  • #8
You don't evaluate any integrals ... and secondly what you have listed is not the bilinear form. The bilinear form is an operator on two functions B(u,v)= integral (_____) ... you don't actually evaluate anything. All you need to show is that B(u,v)=B(v,u) and B is coercive .. try reading up a little on the topic , then work on the problem
 
  • #9
Sorry.
What I was meant to ask is how do I apply the divergence theorem to that integral above.
 
  • #10
You don't use divergence theorem on the integral above. You need to first multiply by a test function, v, then integrate by parts. When you integrate by parts, the laplacian term will actually decompose into two terms. One will be gradient of u times gradient of v and the other will involve divergence. You can then use the divergence theorem on the divergence term to get an integral on the boundary, then you use boundary conditions
 
  • #11
In my lecture notes it says to use the divergence theorem on [tex]\int_D[/tex] (-[tex]\nabla[/tex] . [tex]\nabla[/tex]u + b . [tex]\nabla[/tex]u)v dx.
Do I get [tex]\int_D[/tex] -[tex]\nabla[/tex]u . [tex]\nabla[/tex]v dx - [tex]\int_d[/tex] v[tex]\partial[/tex]u/[tex]\partial\nu[/tex] d(sigma). (d is the boundary)
 
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1. What is a bilinear form and how is it related to linear functionals?

A bilinear form is a mathematical function that takes in two vectors and produces a scalar. It is related to linear functionals because a linear functional is a linear map from a vector space to its underlying field, which can be represented as a bilinear form.

2. What does it mean for a bilinear form to be symmetric?

A bilinear form is symmetric if it produces the same value when the order of the vectors is swapped. In other words, if we switch the order of the inputs, the output remains the same. Mathematically, this can be represented as B(x,y) = B(y,x) for all vectors x and y.

3. How is coercivity related to bilinear forms and linear functionals?

Coercivity is a property of a bilinear form or linear functional where the value of the form or functional is always positive, except when the input vector is zero. It is related to bilinear forms and linear functionals because it helps establish the existence and uniqueness of solutions to certain mathematical problems.

4. Can you give an example of a symmetric and coercive bilinear form?

One example of a symmetric and coercive bilinear form is the dot product between two vectors in Euclidean space. It is symmetric because the order of the vectors does not matter, and coercive because the value is always positive except when the vectors are both zero.

5. How are bilinear forms and linear functionals used in practical applications?

Bilinear forms and linear functionals are used in a variety of practical applications, including physics, engineering, and economics. They can be used to model and solve optimization problems, calculate work and energy in physical systems, and analyze economic systems. They are also important in numerical analysis and computational methods for solving differential equations and other mathematical problems.

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